# Restriction on variable of these : square root of 4-x square root of 8x+4

### 1 Answer | Add Yours

You need to consider the following restriction on what is inside the square root, thus, since you cannot allow a negative value inside the square root `sqrt(8x+4)` , yields:

`8x + 4 >= 0 `

You need to solve the restriction inequality above to evaluate the values of x that make possible the existence of `sqrt(8x+4)` , such that:

`8x >= -4 => x >= -4/8 => x >= -1/2 => x in [-1/2,oo)`

**Hence, the square root `sqrt(8x+4)` can be evaluated for all real `x` numbers `x>=-1/2.` **

Evaluating the restriction on radical `sqrt(4-x)` yields:

`4-x >= 0 => -x >= -4 => x <= 4 =>` `x in (-oo,4]`

**Hence, the square rootÂ `sqrt(4-x)` can be evaluated for all real x numbers x in (-oo,4].**